Hi All,
Please refer to the earlier Post. But here it is again.
Below is the data for a graph. x is the x axis, and Data the applicable Data
for the graph.
X Data formula
1.0000 0.0688 -0.0012
2.0000 0.1277 0.0370
3.0000 0.1800 0.0756
4.0000 0.2272 0.1146
5.0000 0.4471 0.1541
6.0000 0.6924 0.1940
7.0000 1.0000 0.2343
8.0000 1.3455 0.2751
9.0000 1.5503 0.3163
10.0000 1.7955 0.3580
11.0000 2.0820 0.4001

A Polinomial trendline is then applied to the graph line.
A polinmial graph has r2=.9894, so pretty good fit.
The trend formula is y = 0.0148x2 + 0.0375x - 0.0398
This is entered into the Formula column as
((0.0148*x)^2)+(0.0375*x)-0.0398

If this data is taken and put into the same graph as a new line it bears no
remblance to the original polinomial trend line.
Which part of how thick I am, is failing to grasp the simple answer to this?
Please be polite in replies, (though I may be too thick to understand them)
:)

Thanks

Please see the response(s) to your other post.

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article , says...
Hi All,
Please refer to the earlier Post. But here it is again.
Below is the data for a graph. x is the x axis, and Data the applicable Data
{snip}

Hi Hugh,

I can't work out what it is you have done wrong.
If your first X value is in A1 then the formula should be,

A1: =((0.0148*(A1*A1))+(0.0375*A1)-0.0398)

which returns 0.0125

This formula the dragged down produces a plot that is very close to the
trend line.

Cheers
Andy

Hugh wrote:
Hi All,
Please refer to the earlier Post. But here it is again.
Below is the data for a graph. x is the x axis, and Data the applicable Data
for the graph.
X Data formula
1.0000 0.0688 -0.0012
2.0000 0.1277 0.0370
3.0000 0.1800 0.0756
4.0000 0.2272 0.1146
5.0000 0.4471 0.1541
6.0000 0.6924 0.1940
7.0000 1.0000 0.2343
8.0000 1.3455 0.2751
9.0000 1.5503 0.3163
10.0000 1.7955 0.3580
11.0000 2.0820 0.4001

A Polinomial trendline is then applied to the graph line.
A polinmial graph has r2=.9894, so pretty good fit.
The trend formula is y = 0.0148x2 + 0.0375x - 0.0398
This is entered into the Formula column as
((0.0148*x)^2)+(0.0375*x)-0.0398

If this data is taken and put into the same graph as a new line it bears no
remblance to the original polinomial trend line.
Which part of how thick I am, is failing to grasp the simple answer to this?
Please be polite in replies, (though I may be too thick to understand them)
:)

Thanks




--

Andy Pope, Microsoft MVP - Excel
http://www.andypope.info

Many Thanks,Hugh
"Tushar Mehta" wrote in message
...
Please see the response(s) to your other post.

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article , says...
Hi All,
Please refer to the earlier Post. But here it is again.
Below is the data for a graph. x is the x axis, and Data the applicable
Data
{snip}

Many thanks, I appreciate the help.

Hugh
"Andy Pope" wrote in message
...
Hi Hugh,

I can't work out what it is you have done wrong.
If your first X value is in A1 then the formula should be,

A1: =((0.0148*(A1*A1))+(0.0375*A1)-0.0398)

which returns 0.0125

This formula the dragged down produces a plot that is very close to the
trend line.

Cheers
Andy

Hugh wrote:
Hi All,
Please refer to the earlier Post. But here it is again.
Below is the data for a graph. x is the x axis, and Data the applicable
Data for the graph.
X Data formula
1.0000 0.0688 -0.0012
2.0000 0.1277 0.0370
3.0000 0.1800 0.0756
4.0000 0.2272 0.1146
5.0000 0.4471 0.1541
6.0000 0.6924 0.1940
7.0000 1.0000 0.2343
8.0000 1.3455 0.2751
9.0000 1.5503 0.3163
10.0000 1.7955 0.3580
11.0000 2.0820 0.4001

A Polinomial trendline is then applied to the graph line.
A polinmial graph has r2=.9894, so pretty good fit.
The trend formula is y = 0.0148x2 + 0.0375x - 0.0398
This is entered into the Formula column as
((0.0148*x)^2)+(0.0375*x)-0.0398

If this data is taken and put into the same graph as a new line it bears
no remblance to the original polinomial trend line.
Which part of how thick I am, is failing to grasp the simple answer to
this?
Please be polite in replies, (though I may be too thick to understand
them) :)

Thanks




--

Andy Pope, Microsoft MVP - Excel
http://www.andypope.info